Ncert Exercises & Solutions

The time required for 10% completion of a first-order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 10¹⁰s^–1. Calculate k at 318K and Ea.

For a first order reaction,

t = \frac{2.303}{k} \log \frac{a}{a - x}

t = \frac{2.303}{k} \log \frac{100}{90}

A t 298 K,

= \frac{0.1054}{k}

t' = \frac{2.303}{k'} \log \frac{100}{75}

A t 308 K,

= \frac{2.2877}{k'}

According to the question,

t = t'

\Rightarrow \frac{0.1054}{k} = \frac{0.2877}{k'}

\Rightarrow \frac{k'}{k} = 2.7296

From Arrhenius equation, we obtain

\begin{array}{l} \log \frac{k^{'}}{k} = \frac{E_{a}}{2.303 R} (\frac{T^{'} - T}{T T^{'}}) \\ \log (2.7296) = \frac{E_{a}}{2.303 \times 8.314} (\frac{308 - 298}{298 \times 308}) \\ E_{a} = \frac{2.303 \times 8.314 \times 298 \times 308 \times \log (2.7296)}{308 - 298} \\ = 76640.096 {Jmol}^{- 1} \\ = 76.64 kJ {mol}^{- 1} \end{array}

To calculate k at 318 K,

It is given that,

A = 4 \times 10^{10} s^{- 1}, T = 318 K

Again, from the Arrhenius equation, we obtain

\begin{aligned} \log k & = \log A - \frac{E_{a}}{2.303 R T} \\ = \log (4 \times 10^{10}) - \frac{76.64 \times 10^{3}}{2.303 \times 8.314 \times 318} \\ = (0.6021 + 10) - 12.5876 \\ = - 1.9855 \\ Therefore, k & = Antilog (- 1.9855) \\ = 1.034 \times 10^{- 2} s^{- 1} \end{aligned}

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